Mathematics - Matrices Question with Solution | TestHub

$x^2+x+1=0$
$\alpha$ is root
$\begin{array}{rl}& \therefore {\alpha }^2+\alpha +1=0\\ & \Rightarrow \alpha =\omega \text{ as }{\omega }^2\text{ [cube root of unity] }\end{array}$
also
$\begin{array}{l}[\begin{array}{ll}4-a-2b& 64-a-14b\end{array}52+2a-8b\end{array}]$
$\begin{array}{rl}& =\left[\begin{array}{llc}0& 0& 0\end{array}\right]\\ \therefore & a+2b=4\\ & a+14b=64\\ \Rightarrow & 12b=60\Rightarrow b=5\\ \Rightarrow & a=-6\\ \therefore & \frac{4}{{\alpha }^4}+\frac{m}{{\alpha }^{-6}}+\frac{n}{{\alpha }^5}=3\\ \Rightarrow & \frac{4}{\omega }+\frac{m}{1}+\frac{n}{{\omega }^2}=3\\ \Rightarrow & 4{\omega }^2+m+n\omega =3\end{array}$
$\begin{array}{rl}& \Rightarrow 4\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i}\right)+\mathrm{m}+\mathrm{n}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i}\right)=3\\ & \therefore -2+\mathrm{m}-\frac{\mathrm{n}}{2}=\mathrm{3....}(1)\\ & \&\frac{-4\sqrt{3}}{2}+\frac{\mathrm{n}\sqrt{3}}{2}=0\\ & \therefore \mathrm{n}=4\\ & \mathrm{m}=7\\ & \therefore \mathrm{m}+\mathrm{n}=11\end{array}$